The volume of `O_(2)` at STP required for the complete combustion of 4 g `CH_(4)` is

Options

A. 5.6 litre
B. 2.88 litre
C. 22.4 litre
D. 11.2 litre

Answer

To find the volume of \( O_2 \) at STP required for the complete combustion of 4 g of \( CH_4 \), we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of methane (\( CH_4 \)).
The balanced equation is:
\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \]
This indicates that 1 mole of \( CH_4 \) reacts with 2 moles of \( O_2 \).

### Step 2: Calculate the molar mass of \( CH_4 \).
The molar mass of \( CH_4 \) (methane) is calculated as follows:
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol × 4 = 4 g/mol
\[ \text{Molar mass of } CH_4 = 12 + 4 = 16 \text{ g/mol} \]

### Step 3: Determine the number of moles of \( CH_4 \) in 4 g.
Using the molar mass calculated:
\[ \text{Number of moles of } CH_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \text{ g}}{16 \text{ g/mol}} = 0.25 \text{ moles} \]

### Step 4: Use the stoichiometry of the reaction to find the moles of \( O_2 \) required.
From the balanced equation, 1 mole of \( CH_4 \) requires 2 moles of \( O_2 \). Therefore, for 0.25 moles of \( CH_4 \):
\[ \text{Moles of } O_2 = 0.25 \text{ moles } CH_4 \times 2 = 0.5 \text{ moles } O_2 \]

### Step 5: Convert moles of \( O_2 \) to volume at STP.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore:
\[ \text{Volume of } O_2 = 0.5 \text{ moles} \times 22.4 \text{ L/mole} = 11.2 \text{ L} \]

### Conclusion
The volume of \( O_2 \) required for the complete combustion of 4 g of \( CH_4 \) is **11.2 liters**.

---